The first step in responding to any sports injury is to __________.

Physics

2 answers:

Bogdan [553]3 years ago

7 0

Answer:

It would be A.

Explanation:

Ensuring the airway is not closed is second, and some times you may not need CPR, and you should already have told them to stop.

larisa86 [58]3 years ago

6 0

Answer:

THE ANSWER IS..................................................... UHHHH IM JK ITS (A)

Explanation:

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MAVERICK [17]

Answer:

Explanation:

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4 0

3 years ago

The conversion of potential energy in coal into the kinetic energy that heats steam is an expression of: Answer the first law of

erastova [34]

Let me think about this........ correct me if I am wrong....... Laws of thermodynamics 1) Energy cannot be created nor destroyed. It is transferred. 2) All spontaneous reactions will increased the entropy of the universe (More disorder) 3) The entropy of the crystal lattice structure at 0 K is zero PE(coal) --> Steam(K.E) ................ This will increase entropy and also conversion of energy. So BOTH LAWS is my final answer. Wait...... I would like to use a lifelines...... I want to phone my Physical chem professor. *Ring Ring.........* Me: Hello, I wanted to make sure if I got the correct answer. Chem prof: You dont need my help, since you got the right answer. Me: WOOT!!!!!!!

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3 years ago

A(n) 1700 kg car is moving along a level road at 21 m/s. The driver accelerates, and in the next 10 s the engine provides 22000

allochka39001 [22]

The final speed of the car at the given conditions is 30.1 m/s.

The given parameters:

Mass of the car, m = 1700 kg
Velocity of the car, v = 21 m/s
Time of motion, t = 10 s
Additional energy provided by the engine, E₁ = 22,000 J
Energy used in overcoming friction, E₂ = 3,666.67 J

The change in the energy applied to the car is calculated as;

$\Delta E = E_1 - E_2\\\\\Delta E = 22,000 \ J \ - \ 3,666.67 \ J\\\\\Delta E = 18,333.33 \ J$

The final speed of the car is calculated as follows;

$\Delta E = \frac{1}{2} m(v_f^2 - v_0^2)\\\\v_f^2 - v_0^2 = \frac{2\Delta E}{m} \\\\v_f^2 = \frac{2\Delta E}{m} + v_0^2\\\\v_f = \sqrt{ \frac{2\Delta E}{m} + v_0^2} \\\\v_f = \sqrt{ \frac{2\times 18,333.4}{1700} + (21)^2} \\\\v_f = 30.1 \ m/s$